Section 10.4: Motion in Space: Velocity and Acceleration

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Section 10.4: Motion in Space: Velocity and Acceleration

Practice HW from Stewart Textbook (not to hand in)

p. 725 # 3-17 odd, 21, 23

Velocity and Acceleration

Given a vector function r(t ) = f (t) i + g (t) j + h (t) k , we define the following

quantities:

Velocity = v(t ) = r ¡ä(t ) = f ¡ä(t ) i + g ¡ä(t ) j + h¡ä(t ) k

Speed = | v (t ) |= | r ¡ä(t ) |= [ f ¡ä(t )] 2 + [ g ¡ä(t )]2 + [h ¡ä(t )] 2

Acceleration = a(t ) = v ¡ä(t ) = r ¡ä¡ä(t ) = f ¡ä¡ä(t ) i + g ¡ä¡ä(t ) j + h¡ä¡ä(t ) k

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Example 1: Find the velocity, acceleration, and speed of a particle given by the position

function r (t ) =< t + 1, t 2 > at t = 2. Sketch the path of the particle and draw the velocity

and acceleration vectors for the specified value of t.

.

Solution:

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Example 2: Find the velocity, acceleration, and speed of a particle given by the position

function r (t ) = 2 cos t i + 3 sin t j at t = 0. Sketch the path of the particle and draw the

velocity and acceleration vectors for the specified value of t.

.

Solution: We first calculate the velocity, speed, and acceleration formulas for an

arbitrary value of t. In the process, we substitute and find each of these vectors at t = 0.

Velocity = v(t ) = r ¡ä(t ) = ?2 sin t i + 3 cos t j

Velocity at

t=0

= v (0) = ?2 sin 0 i + 3 cos 0 j = ?2(0 _ i + 3(1) j = 3 j =< 0,3 >

Speed =| v (t ) |= (?2 sin t ) 2 + (3 cos t ) 2 = 4 sin 2 t + 9 cos 2 t

Speed at

= | v (0) |= 4 sin 2 0 + 9 cos 2 0 = 4(0) 2 + 9(1) 2 = 0 + 9 = 9 = 3

t=0

Acceleration = a(t ) = v ¡ä(t ) = ?2 cos t i ? 3 sin t j

Acceleration at

= a(0) = ?2 cos 0 i ? 3 sin 0 j = ?2(1)i ? 3(0) j = ?2i =< ?2,0 >

t=0

To graph the path of the particle, we take the position r (t ) = 2 cos t i + 3 sin t j , write the

parametric equations x = 2 cos t and y = 3 sin t , and solve for the trigonometric terms to

x

y

get cos t = and sin t = . Rewriting the Pythagorean identity cos 2 t + sin 2 t = 1 as

2

3

(cos t ) 2 + (sin t ) 2 = 1

and substituting cos t =

x

y

and sin t = , we obtain

2

3

2

2

? x?

? y?

? ? +? ? =1

? 2?

? 3?

or

x2 y2

+

=1

4

9

This is the equation of an ellipse with center (0, 0). The vertices on the x-axis are (2, 0)

and (-2, 0) formed by taking a = 4 = 2 and moving to the left and right of the center on

(continued on next page)

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the x-axis. The vertices on the y-axis are (0, 3) and (0, -3) formed by taking b = 9 = 3

and moving up and down from the center on the y-axis. To find the position of the

particle at time t = 0, we take the position function r (t ) = 2 cos t i + 3 sin t j and substitute

t = 0 to obtain

r (0) = 2 cos 0 i + 3 sin 0 j = 2(1) i + 3(0) j = 2 i =< 2,0 > .

The terminal point of this vector, (2, 0), gives the initial point to plot the velocity and

acceleration vectors at t = 0 we found above, v (0) =< 0,3 > and a(0) =< ?2,0 > . The

x2 y2

+

=1

4

9

plotted in green, the velocity vector v = < 0, 3 > in blue, and the acceleration vector

a = < -2, 0 > in red.

following shows the result of the graph, with the equation of the ellipse

v =< 0, 3 >

a =< -2, 0 >

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Example 3: Find the velocity, acceleration, and speed of a particle given by the position

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function r (t ) = e t i + sin 4t j + t 2 k

Solution:

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